Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{k^2 - 5k}{10k - 10} \times \dfrac{k - 1}{k^2 - 3k - 10} $
Explanation: First factor the quadratic. $z = \dfrac{k^2 - 5k}{10k - 10} \times \dfrac{k - 1}{(k - 5)(k + 2)} $ Then factor out any other terms. $z = \dfrac{k(k - 5)}{10(k - 1)} \times \dfrac{k - 1}{(k - 5)(k + 2)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ k(k - 5) \times (k - 1) } { 10(k - 1) \times (k - 5)(k + 2) } $ $z = \dfrac{ k(k - 5)(k - 1)}{ 10(k - 1)(k - 5)(k + 2)} $ Notice that $(k - 1)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ k\cancel{(k - 5)}(k - 1)}{ 10(k - 1)\cancel{(k - 5)}(k + 2)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $z = \dfrac{ k\cancel{(k - 5)}\cancel{(k - 1)}}{ 10\cancel{(k - 1)}\cancel{(k - 5)}(k + 2)} $ We are dividing by $k - 1$ , so $k - 1 \neq 0$ Therefore, $k \neq 1$ $z = \dfrac{k}{10(k + 2)} ; \space k \neq 5 ; \space k \neq 1 $